Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{8r}{18r^2 + 12r} \times \dfrac{9r + 6}{2} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ 8r \times (9r + 6) } { (18r^2 + 12r) \times 2 } $ $ n = \dfrac {8r \times 3(3r + 2)} {2 \times 6r(3r + 2)} $ $ n = \dfrac{24r(3r + 2)}{12r(3r + 2)} $ We can cancel the $3r + 2$ so long as $3r + 2 \neq 0$ Therefore $r \neq -\dfrac{2}{3}$ $n = \dfrac{24r \cancel{(3r + 2})}{12r \cancel{(3r + 2)}} = \dfrac{24r}{12r} = 2 $